Can a non-constant analytic function have infinitely many zeros on a closed disk?

Solution 1:

The relevance of the fact that the set is closed is that the limit point must be in the set. For example $\sin (1/(z+1))$ has infinitely many zeros in the open unit disc, but is not zero. (The sole point of accumulation is $-1$, outside the domain, and the function is not analytic there.)

On the multiplicity: first I think distinct zeros are meant. Second, a non-zero analytic function (on a connected domain) cannot vanish to infinite order anywhere; this would give the series there is $0$ and so the function is zero.

Solution 2:

Since the disk $\mathcal{D}$ is bounded and closed it is compact, so, by Heine-Borel it fulfils the Bolzano-Weierstrass property: every infinite set has at least one limit point. So the set of $\mathcal{Z}$ of zeros has a limit point, and you're right to go with the identity theorem. So you can see the reason for closure: it lets you call on the Heine-Borel theorem.

Incidentally, before about 1930 and Pavel Alexandrof's school, "compact" meant by definition having the Bolzano-Weierstrass property.